Algebra stuff...again!?!?

Question

the problem is (2x-1)^2=25 but i can't solve for x

i know x=3 but i can't figure out how to show my work

the furthest i can get is 4x^2-4x+1=25

please help!!!

Answer 1

(2x-1)(2x-1)=25

4x^2-2x-2x=1=25

4x^2-4x+24=0

divide it all by four

x^2-x+6=0

(x+2)(x-3)

x=-2 or 3

Answer 2

25 is a perfect square, so take the square root of both sides.  Since the square root can be positive or negative, you'll have two equations to solve:

2x - 1 = 5  and 2x - 1 = -5

Solve each of them and you'll have your answers.  x = 3 is one, but there's another.

Answer 3

holy c**p. I'm starting algebra in a week. this is what I'm gonna have to do... ok i believe you reverse the steps (you know the PEMDAS) to get the answer.

ps - sorry if I'm wrong. i haven't started school yet

Answer 4

(2x-1)^2 = 25

You can factor out (2x-1)^2 but since 25 is a perfect square, you can simply remove the square on the binomial.

1ST APPROACH

2x-1 = +/- sqrt(25)   <-- remember that a number has two square roots (+,-)

2x-1 = +/- 5

Solve for x:

2x - 1 = 5

2x = 6

x=3

2x - 1 = -5

2x = -4

x = -2

.: x = 3, -2

----

2ND APPROACH

(2x-1)^2 = 25

4x^2 - 4x + 1 = 25  [use FOIL or remember that (a+b)^2 = a^2+2ab+b^2 ]

4x^2 - 4x - 24 = 0   <--transpose 25

x^2 - x - 6 = 0   <--divide both sides by 4

appch2.1

use the quadratic formula: x = [-b +/- sqrt(b^2-4ac)] / 2a

a = 1, b = -1, c = -6

x = [1 +/- sqrt(1+24)] / 2

x = [1+/-5]/2

.: x = 3, -2

appch2.2

complete the square:

x^2 - x -6 = 0

x^2 - x = 6

x^2 - x (-1/2)^2 = 6 + (-1/2)^2

(x-1/2)^2 = 25/4

Still, you'll arrive at x = 3, -2